11.4 Colligative properties  (Page 8/30)

Determination of molar masses

Osmotic pressure and changes in freezing point, boiling point, and vapor pressure are directly proportional to the concentration of solute present. Consequently, we can use a measurement of one of these properties to determine the molar mass of the solute from the measurements.

Determination of a molar mass from a freezing point depression

A solution of 4.00 g of a nonelectrolyte dissolved in 55.0 g of benzene is found to freeze at 2.32 °C. What is the molar mass of this compound?

Solution

We can solve this problem using the following steps.

1. Determine the change in freezing point from the observed freezing point and the freezing point of pure benzene ( [link] ).
   
   $\text{Δ}{T}_{\text{f}}=5.5\text{°}\text{C}-2.32\text{°}\text{C}=3.2\text{°}\text{C}$
2. Determine the molal concentration from K _f , the freezing point depression constant for benzene ( [link] ), and Δ T _f .
   
   $\begin{array}{l}\hfill \text{Δ}{T}_{\text{f}}=K_{\text{f}}m\hfill \\ \\ m=\frac{\text{Δ}{T}_{\text{f}}}{K_{\text{f}}}=\frac{3.2\text{°}\text{C}}{5.12\text{°}\text{C}{m}^{\mathrm{-1}}}=0.63m\end{array}$
3. Determine the number of moles of compound in the solution from the molal concentration and the mass of solvent used to make the solution.
   
   $\text{Moles of solute}=\frac{0.62\text{mol solute}}{1.00\sout{\text{kg solvent}}}\times 0.0550\sout{\text{kg solvent}}=0.035\text{mol}$
4. Determine the molar mass from the mass of the solute and the number of moles in that mass.
   
   $\text{Molar mass}=\frac{4.00\text{g}}{0.034\text{mol}}=1.2\times {10}^2\text{g/mol}$

Check your learning

A solution of 35.7 g of a nonelectrolyte in 220.0 g of chloroform has a boiling point of 64.5 °C. What is the molar mass of this compound?

Answer:

1.8 $\times$ 10 ² g/mol

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Determination of a molar mass from osmotic pressure

A 0.500 L sample of an aqueous solution containing 10.0 g of hemoglobin has an osmotic pressure of 5.9 torr at 22 °C. What is the molar mass of hemoglobin?

Solution

Here is one set of steps that can be used to solve the problem:

1. Convert the osmotic pressure to atmospheres, then determine the molar concentration from the osmotic pressure.
   
   $\begin{array}{}\\ \Pi =\frac{5.9\text{torr}\times 1\text{atm}}{760\text{torr}}=7.8\times {10}^{\mathrm{-3}}\text{atm}\\ \Pi =\mathit{\text{MRT}}\\ \\ M=\frac{\Pi }{RT}=\frac{7.8\times {10}^{\mathrm{-3}}\text{atm}}{\left(0.08206\text{L atm/mol K}\right)\left(295\text{K}\right)}=3.2\times {10}^{\mathrm{-4}}\text{M}\end{array}$
2. Determine the number of moles of hemoglobin in the solution from the concentration and the volume of the solution.
   
   $\text{moles of hemoglobin}=\frac{3.2\times {10}^{\mathrm{-4}}\text{mol}}{1\sout{\text{L solution}}}\times 0.500\sout{\text{L solution}}=1.6\times {10}^{\mathrm{-4}}\text{mol}$
3. Determine the molar mass from the mass of hemoglobin and the number of moles in that mass.
   
   $\text{molar mass}=\frac{10.0\text{g}}{1.6\times {10}^{\mathrm{-4}}\text{mol}}=6.2\times {10}^4\text{g/mol}$

Check your learning

What is the molar mass of a protein if a solution of 0.02 g of the protein in 25.0 mL of solution has an osmotic pressure of 0.56 torr at 25 °C?

Answer:

2.7 $\times$ 10 ⁴ g/mol

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Colligative properties of electrolytes

As noted previously in this module, the colligative properties of a solution depend only on the number, not on the kind, of solute species dissolved. For example, 1 mole of any nonelectrolyte dissolved in 1 kilogram of solvent produces the same lowering of the freezing point as does 1 mole of any other nonelectrolyte. However, 1 mole of sodium chloride (an electrolyte) forms 2 moles of ions when dissolved in solution. Each individual ion produces the same effect on the freezing point as a single molecule does.